Question

A uniform bar of length 100 cm and weight 50 N has two points on the bar such that BC = 20 cm and BD = 30 cm. It is supported horizontally by two vertical strings, one attached at point C and the other at B. A load of weight 30 N is suspended from A, and another load of weight T N is suspended from D. Find the tension in each string.

A. T = 30 N
B. T = 40 N
C. T = 50 N
D. T = 60 N

Answer 1

The tension in string BC is 50 N and the tension in string BD is 80 N.

Step-by-step explanation:

To find the tension in each string, we can start by analyzing the forces acting on the bar. Let's consider the forces on point C. There is a tension force T1 acting upwards and the weight of the bar acting downwards. Since the bar is in equilibrium, the sum of these forces must be zero. Therefore, we have:

T1 - 50 N = 0

T1 = 50 N

Now let's consider the forces on point B. There is a tension force T2 acting upwards and the weight of the bar acting downwards. Additionally, there is a weight force of 30 N acting downwards at point D. Again, since the bar is in equilibrium, we have:

T2 - 50 N - 30 N = 0

T2 = 80 N

Therefore, the tension in string BC is 50 N (Option B) and the tension in string BD is 80 N (Option D).

Answer 2

Using the principles of static equilibrium, we calculate the torques about point B on the bar to find the unknown tension T. However, after applying the formulas, the calculated tension T = 1100 N does not match the options provided, indicating a possible mistake in the question's values or in the interpretation of the question.

Step-by-step explanation:

To find the tension in each string supporting the uniform bar with weights suspended from it, we must apply the principles of static equilibrium, where the sum of the forces and the sum of the torques (moments) are both equal to zero. Let's designate the following points on the bar: A at the far left end, B at 70 cm from A, C at 50 cm from A, and D at 40 cm from A. The weights at A and D are 30 N and T N, respectively. The weight of the bar acts at its center of gravity, which is at the midpoint (at 50 cm from either end).

First, we calculate the torque around point B because the tension at this point does not produce any torque about B. We have:

• The torque due to the weight at A (30 N) is 30 N × 0.7 m (counter-clockwise).
• The torque due to the weight of the bar (50 N) is 50 N × 0.2 m (clockwise) as it acts at 20 cm to the left of B.
• The torque due to the weight at D (T N) is T N × 0.1 m (clockwise).

Now, for static equilibrium, these torques must balance:

30 N × 0.7 m = 50 N × 0.2 m + T N × 0.1 m

21 Nm = 10 Nm + 0.1 T Nm

0.1 T Nm = 21 Nm - 10 Nm

T = 110 N / 0.1 m

T = 1100 N (This is higher than the options provided, suggesting there might be a mistake in the values given in the question, the question itself, or the calculations).

Once T is known, tensions in the strings can be found by the sum of forces in the vertical direction equals zero:

Let Tc be the tension in the string at C and Tb the tension in the string at B, then:

Tc + Tb = weight of bar + weight at A + T

Tc + Tb = 50 N + 30 N + 1100 N

Tc + Tb = 1180 N (Since the question is likely to contain errors, this is a placeholder calculation based on prior incorrect T value)